Question

A cubical block of side 'a' is moving with a velocity 'v' on a smooth horizontal plane as shown in the figure. It hits a ridge at point O. The angular speed of the block after it hits O is:

• A. $\frac{3v}{4a}$
• B. $\frac{3v}{2a}$
• C. $\frac{\sqrt{3}v}{\sqrt{2}a}$
• D. Zero

Answer 1

The correct option is A $\frac{3v}{4a}$

Conserving Angular momentum before and after collision
$mv\times \frac{a}{2}=I\omega$
Now, $I=\frac{m(\sqrt{2}a{)}^2}{12}+m{\left(\frac{a}{\sqrt{2}}\right)}^2$
$=m{a}^2(\frac{1}{6}+\frac{1}{2})$
$=\frac{2}{3}m{a}^2.$
$\therefore \omega =\frac{mva}{2}\times \frac{3}{2m{a}^2}=\frac{3v}{4a}.$